Let A be a real $n \times n$ matrix such that $\langle Ax,x\rangle \geq 0$ Prove that $Au=0$ iff $A^tu=0$.

Question

Let A be a real $n \times n$ matrix such that $\langle Ax,x\rangle \geq 0$ Prove that $Au=0$ iff $A^tu=0$.

Now if we have $Au=0$ for some $u=0$ then clearly $A^tu=0$.

Hence, $Au=0$ for some $u\neq 0$ then on a contrary if $A^tu\neq 0$ then from $\langle A^tu,u \rangle = \langle u,Au \rangle = 0$ do we get any contradiction? Please help from here.

Answer 1

Let $v \in \mathbb R^n$ :

For every $t \in \mathbb R$, $$0 \le \langle A(u + tv), (u+tv)\rangle = \langle A^t (u + tv) , (u + tv)\rangle = \langle A^tu , u\rangle + t\langle A^t v,u\rangle + t \langle A^tu,v\rangle + t^2 \langle A^tv, v\rangle = t\left(\langle A^tu,v\rangle + t \langle A^tv, v\rangle\right)$$ taking $t > 0$ then dividing by $t$ we have $$\langle A^tu,v\rangle + t \langle A^tv, v\rangle \ge 0$$ Now $t\to 0$, $$\langle A^t u ,v \rangle \ge 0$$ taking $t < 0$ then dividing by $t$ we have $$\langle A^tu,v\rangle + t \langle A^tv, v\rangle \le 0$$ Now $t\to 0$, $$\langle A^t u ,v \rangle \le 0$$ So $$\langle A^tu, v\rangle = 0$$ This is true for every $v$ then by taking $v = A^tu$ we have $$\left\|A^tu\right\|^2 = \langle A^t u, A^t u \rangle = 0$$